Have you graphed these? \(\displaystyle y= 4- x^2\) is, of course, a parabola, opening downward, with vertex at (0, 4) and crossing the x-axis at (-2, 0) and (2, 0). \(\displaystyle y= 2- x\) is a straight line with x- intercept (2, 0) and y-intercept (0, 2). That line crosses the parabola where \(\displaystyle 4- x^2= 2- x\) so where \(\displaystyle x^2- x- 2= (x+ 1)(x- 2)= 0\): at (-1, 3) and (2, 0). The line x= -2 is, of course, just where the parabola crosses the x-axis.
The line x= 3 (rather than x= 2) is a little peculiar. It crosses the x-axis well away from the portion of the parabola above the x-axis. The region "bounded by" these four graph has two parts! The first part is from x= -2 to x= 2 where the parabola lies above the line so you need to integrate \(\displaystyle 4- x^2- (2- x)= 2+ x- x^2\) and the second part is from x= 2 to x= 3 where the line lies above the parabola so you need to integrate \(\displaystyle 2- x- (4- x^2)= x^2- x- 2\).
